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Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0

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Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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Is \((x – 2)^2 > x^2\)?


(1) \(x^2 > x\)

(2) \(\frac{1}{x} > 0\)
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 30 Nov 2009, 08:28
kirankp wrote:
Is (x – 2)^2 > x^2?

(A) x^2 > x

(B) 1/x > 0


answer C:

A) x^2 > x

if x = -2 then (-2-2)^2 > -2^2 = 16 > 4 = yes
if x = 2 then (2-2)^2 > 2^2 = 0 > 4 = no

B) 1/x > 0

if x = 2 then answer is no from above
if x = 1/2 then (1/2-2)^2 > 1/2 ^2 = (-3/2)^2 > 1/2 ^2 = 9/4 > 1/4 = yes

together sufficient
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 30 Nov 2009, 08:31
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Is (x – 2)^2 > x^2?
(A) x^2 > x
(B) 1/x > 0

Answer: C

A) is not sufficient by itself as x can be positive or negative, but x cannot be a fraction i.e., x is not between -1 and 1 (inclusive)
the question stem is true when x is negative and false when x is positive so insufficient

B) from this , x must be positive or x>0

when x is 1/2 , question stem is TRUE, when x is >= 1 its false; so B by itself insufficient

taking together we know that x is > 1 and the question stem is always FAlse... hence answer C
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Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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Is (x - 2)^2 > x^2?

Let's work on the stem first:

Is \((x-2)^2>x^2\)? --> Is \(x^2-4x+4>x^2\)? --> Is \(x<1\)? So this is what the question is basically asking.

(1) \(x^2>x\) --> \(x(x-1)>0\) --> \(x<0\) OR \(x>1\). Two ranges, not sufficient.

(2) \(\frac{1}{x}>0\) --> \(x>0\). Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is \(x>1\). Hence the answer to the question is no. Sufficient.

Answer: C.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 08 Dec 2009, 22:29
Bunuel wrote:
Let's work on the stem first:

Is \((x-2)^2>x^2\)? --> Is \(x^2-4x+4>x^2\)? --> Is \(x<1\)? So this is what the question is basically asking.

(1) \(x^2>x\) --> \(x(x-1)>0\) --> \(x<0\) OR \(x>1\). Two ranges, not sufficient.

(2) \(\frac{1}{x}>0\) --> \(x>0\). Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is \(x>1\). Hence the answer to the question is no. Sufficient.

Answer: C.


I could not get two points

1. How did you come to x<0 in statemnt 1? I mean x(x-1)> 0 ==> Doesn't this mean that x>0 & x>1 ??

2. How do we find the intersection of the two results in above two statements??
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 20 Dec 2009, 08:03
Hussain15 wrote:
Bunuel wrote:
Let's work on the stem first:

Is \((x-2)^2>x^2\)? --> Is \(x^2-4x+4>x^2\)? --> Is \(x<1\)? So this is what the question is basically asking.

(1) \(x^2>x\) --> \(x(x-1)>0\) --> \(x<0\) OR \(x>1\). Two ranges, not sufficient.

(2) \(\frac{1}{x}>0\) --> \(x>0\). Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is \(x>1\). Hence the answer to the question is no. Sufficient.

Answer: C.


I could not get two points

1. How did you come to x<0 in statemnt 1? I mean x(x-1)> 0 ==> Doesn't this mean that x>0 & x>1 ??

2. How do we find the intersection of the two results in above two statements??


1. How did you come to x<0 in statemnt 1? I mean x(x-1)> 0 ==> Doesn't this mean that x>0 & x>1 ??
You get three sets of the number : (infinity,0) or (0,1) or (1, infinity)
Then, Just plug in the numbers :
x = -1, then 2 > 0
x = 1/2, then -1/4 < 0
x = 2, then 2 > 0
Thus, you get (infinity,0) or (1, infinity)

2. How do we find the intersection of the two results in above two statements??
stm 1 : x < 0 or x > 1
stm 2 : x > 0
Both : x > 1

Both means that x must be true for the two statements. Thus, only x > 1 fits the definition

Hope it helps
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 17 Mar 2011, 23:23
Bunuel,
how did you get this:
(1)+(2) Intersection of the ranges from (1) and (2) is x>1 so the answer to the question "is x<1?" is NO. Sufficient.
The way I see it, statement 1 says x is less than 0 or x is greater than 1
statement 2 says x is greater than 0.
combining the two statements I see an infinite range.

x<0 AND x>0 AND x>1.
The limiting range here is x<0 AND x>1
How did you get your solution?
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 18 Mar 2011, 00:36
thesfactor wrote:
Bunuel,
how did you get this:
(1)+(2) Intersection of the ranges from (1) and (2) is x>1 so the answer to the question "is x<1?" is NO. Sufficient.
The way I see it, statement 1 says x is less than 0 or x is greater than 1
statement 2 says x is greater than 0.
combining the two statements I see an infinite range.

x<0 AND x>0 AND x>1.
The limiting range here is x<0 AND x>1
How did you get your solution?


1st statement: x<0 OR x>1
2nd statement: x>0

Combining both;
(x<0 OR x>1) AND x>0

2nd statement says; I am telling you that x>0; accept it is as truth.
1st statement says: x could be less than 0 OR x could be more than 1. But, as statement 2 already told us that x>0, we should ignore x<0 of the statement 1.

What remains is; x>0 and x>1(limiting)

Hence; x>1
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 18 Mar 2011, 00:52
Is x^2 -4x + 4 > x^2 ?

or 4-4x > 0

or 1 > x

or x < 1


(1) x^2 > x

This can be true for numbers NOT in the range -1 < x < 1


But not sufficient


2. 1/x > 0 means x > 0

But not sufficient

So From (1) and (2) x > 1, so answer is C.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 25 Mar 2014, 23:37
kirankp wrote:
Is (x – 2)^2 > x^2?

(1) x^2 > x
(2) 1/x > 0


Sol: the given equation can be reduced to 4-4x >0 or 4(1-x) >0 or is x<1

St 1 x^2-x > 0 or x(x-1)>0
So we have either x>0 and x>1 or x<0 or x<1. We have yes and no as possible answers so st1 is not enough

St2 1/x> 0 thus x>0 but then x can be less than 1 or more than 1

Combining we get that x>0 and x > 1 thus ans is C

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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 14 Apr 2014, 02:28
Need some help here.

Is (x - 2)^2 > x^2?

Let's work on the stem first:

Is (x-2)^2>x^2? --> Is x^2-4x+4>x^2? --> Is x<1? So this is what the question is basically asking.

(1) x^2>x --> x(x-1)>0 --> x<0 OR x>1. Two ranges, not sufficient. As per me it should be x>0 and x>1. I know this is wrong as I tried with 1/2 as a value of x which falsifies the equation.

Per my understanding,
if the equation were x(x-1) = 0, x can have two values i.e. 0 & 1. Please help me understand if I am missing out something.

(2) \frac{1}{x}>0 --> x>0. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is x>1. Hence the answer to the question is no. Sufficient.

Answer: C.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 14 Apr 2014, 04:45
honey86 wrote:
Need some help here.

Is (x - 2)^2 > x^2?

Let's work on the stem first:

Is (x-2)^2>x^2? --> Is x^2-4x+4>x^2? --> Is x<1? So this is what the question is basically asking.

(1) x^2>x --> x(x-1)>0 --> x<0 OR x>1. Two ranges, not sufficient. As per me it should be x>0 and x>1. I know this is wrong as I tried with 1/2 as a value of x which falsifies the equation.

Per my understanding,
if the equation were x(x-1) = 0, x can have two values i.e. 0 & 1. Please help me understand if I am missing out something.

(2) \frac{1}{x}>0 --> x>0. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is x>1. Hence the answer to the question is no. Sufficient.

Answer: C.


x>0 and x>1 does not make any sense.

x(x-1)>0 --> the "roots" are 0 and 1, this gives us 3 ranges:

x<0;
0<x<1;
x>1.

Next, test some extreme value for x: if x is some large enough number, say 10, then both multiples will be positive which gives the positive result for the whole expression, so when x>1 the expression is positive. Now the trick: as in the 3rd range expression is positive then in the 2nd it'll be negative and finally in the 1st it'll be positive again: + - + . So, the ranges when the expression is positive are: x<0 and x>1.

Check this for more: x2-4x-94661.html#p731476

Hope it helps.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 14 Apr 2014, 05:22
Got it!
Thanks a lot Bunuel. You have explained it very well on the other link that you posted.

Will work on more such problems.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 03 May 2014, 07:45
How about this case?
Since its not mentioned x an is integer or a fraction, let x = 3/2
So stmt becomes (3/2 - 2)^2 > (3/2) which is false.

Ans C is when 0 < x < 1
But when x > 1, then ans should be E
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 05 May 2014, 02:47
Sailesh1986 wrote:
How about this case?
Since its not mentioned x an is integer or a fraction, let x = 3/2
So stmt becomes (3/2 - 2)^2 > (3/2) which is false.

Ans C is when 0 < x < 1
But when x > 1, then ans should be E


x cannot be less than or equal to 1, because when we combine the statements we get that x>1.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 13 Mar 2016, 05:57
Nice Question...
Here all statement 1 gives the range as X<0 or X>1 => not sufficient
Statement 2 gives the range X>0 => Not sufficient
Combining the statements => range = X>1
so w get a NO as a answer => x is never less then zero
Sufficient => C is sufficient
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 18 Oct 2016, 07:54
stonecold wrote:
Nice Question...
Here all statement 1 gives the range as X<0 or X>1 => not sufficient
Statement 2 gives the range X>0 => Not sufficient
Combining the statements => range = X>1
so w get a NO as a answer => x is never less then zero
Sufficient => C is sufficient



Agree: After some solving we find that Q asks whether x<1
S1: X<0 or X>1 ; for 2 answer is NO; for -2 answer is YES
S2: X>0 ; for 2 answer is NO; for 0.5 answer is YES
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post Updated on: 27 Jan 2018, 07:57
Why not A?

In 1, if x^2>x, then wouldn't x>1?

Originally posted by urvashis09 on 27 Jan 2018, 06:46.
Last edited by urvashis09 on 27 Jan 2018, 07:57, edited 1 time in total.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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tamal99 wrote:
Is (x – 2)^2 > x^2?

(1) x^2 > x
(2) 1/x > 0


\((x – 2)^2 > x^2.....x^2-4x+4>x^2....4x<4....x<1\)
So Q basically asks us - Is x<1?

(1) \(x^2 > x\)..
\(x^2 > x....x^2-x>0...x(x-1)>0\)..
so if x>0, x-1>0 or x>1...NO
if x<0, x-1<0 or x<1...YES
insuff

(2) \(\frac{1}{x} > 0\)..
\(\frac{1}{x} > 0\)..
this tells us that x >0
insuff

combined
x>0, so x>1
ans is NO
suff
C
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]

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New post 29 Jan 2018, 00:11
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urvashis09 wrote:
Why not A?

In 1, if x^2>x, then wouldn't x>1?


Hi Urvashi

If x^2 > x , then it could mean two things:
Either x > 1 Or
x < 0 (x can take any negative value which would make its square positive, positive is always greater than negative).
Thats why first statement alone is NOT sufficient.
Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0   [#permalink] 29 Jan 2018, 00:11
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